Question

Given that y(t)=c1e4t+c2e−4ty(t)=c1e4t+c2e−4t is a solution to the differential equation y′′−16y=0y′′−16y=0, where c1c1 and c2c2 are arbitrary constants, find a function y(t)y(t) that satisfies the conditions

Answer 1

Solution :- Given differential equation

$y′′−16y=0$

 So the characteristic equation will be

$r^{2} -16=0 ⇒r^{2} =16 ⇒[tex]r_{1}=4 \\or \\ r_{2}=-4$

so, the  required  function will be

$y=c_{1}e^{r_{1}t}+c_{1}e^{r_{2}t}$

⇒ $y=c_{1}e^{4t}+c_{1}e^{-4t}$ ia solution of the  given differential equation  

where $c_{1} \\and \\c_{2}$ are constants.